【刷题】BZOJ 2816 [ZJOI2012]网络

Description

http://www.lydsy.com/JudgeOnline/upload/zjoi2012.pdf

Solution

维护树上联通块的信息，支持动态加边删边
LCT
总共只有10种颜色，直接建10个LCT，每个LCT维护一种颜色
LCT还是差不多
只是第二个操作比较麻烦，得一个一个颜色地去试

#include<bits/stdc++.h>
#define ll long long
#define db double
#define ld long double
#define lc(x) ch[(x)][0]
#define rc(x) ch[(x)][1]
const int MAXN=10000+10,MAXC=15;
int n,m,c,k,Val[MAXN],d[MAXC][MAXN];
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
struct LCT{
	int ch[MAXN][2],fa[MAXN],rev[MAXN],Mx[MAXN];
	inline void init()
	{
		memset(ch,0,sizeof(ch));
		memset(fa,0,sizeof(fa));
		memset(rev,0,sizeof(rev));
		memset(Mx,0,sizeof(Mx));
	}
	inline bool nroot(int x)
	{
		return lc(fa[x])==x||rc(fa[x])==x;
	}
	inline void reverse(int x)
	{
		std::swap(lc(x),rc(x));
		rev[x]^=1;
	}
	inline void pushup(int x)
	{
		Mx[x]=Val[x];
		chkmax(Mx[x],Mx[lc(x)]);
		chkmax(Mx[x],Mx[rc(x)]);
	}
	inline void pushdown(int x)
	{
		if(rev[x])
		{
			if(lc(x))reverse(lc(x));
			if(rc(x))reverse(rc(x));
			rev[x]=0;
		}
	}
	inline void rotate(int x)
	{
		int f=fa[x],p=fa[f],c=(rc(f)==x);
		if(nroot(f))ch[p][rc(p)==f]=x;
		fa[ch[f][c]=ch[x][c^1]]=f;
		fa[ch[x][c^1]=f]=x;
		fa[x]=p;
		pushup(f);
		pushup(x);
	}
	inline void splay(int x)
	{
		std::stack<int> s;
		s.push(x);
		for(register int i=x;nroot(i);i=fa[i])s.push(fa[i]);
		while(!s.empty())pushdown(s.top()),s.pop();
		for(register int y=fa[x];nroot(x);rotate(x),y=fa[x])
			if(nroot(y))rotate((lc(y)==x)==(lc(fa[y])==y)?y:x);
		pushup(x);
	}
	inline void access(int x)
	{
		for(register int y=0;x;x=fa[y=x])splay(x),rc(x)=y,pushup(x);
	}
	inline void makeroot(int x)
	{
		access(x);splay(x);reverse(x);
	}
	inline int findroot(int x)
	{
		access(x);splay(x);
		while(lc(x))pushdown(x),x=lc(x);
		splay(x);
		return x;
	}
	inline void split(int x,int y)
	{
		makeroot(x);access(y);splay(y);
	}
	inline void link(int x,int y)
	{
		makeroot(x);
		if(findroot(y)!=x)fa[x]=y;
	}
	inline void cut(int x,int y)
	{
		makeroot(x);
		if(findroot(y)==x&&fa[y]==x&&!lc(y))rc(x)=fa[y]=0,pushup(x);
	}
};
LCT T[MAXC];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char c='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(c!='\0')putchar(c);
}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
int main()
{
	read(n);read(m);read(c);read(k);
	for(register int i=1;i<=c;++i)T[i].init();
	for(register int i=1;i<=n;++i)read(Val[i]);
	for(register int i=1;i<=m;++i)
	{
		int u,v,w;
		read(u);read(v);read(w);
		w++;
		d[w][u]++;d[w][v]++;
		T[w].link(u,v);
	}
	while(k--)
	{
		int opt;
		read(opt);
		if(opt==0)
		{
			int x,y;
			read(x);read(y);
			Val[x]=y;
			for(register int i=1;i<=c;++i)T[i].access(x),T[i].splay(x),T[i].pushup(x);
		}
		if(opt==1)
		{
			int u,v,w,dn=1;
			read(u);read(v);read(w);
			w++;
			for(register int i=1;i<=c;++i)
				if(T[i].findroot(u)==T[i].findroot(v))
				{
					T[i].makeroot(u),T[i].access(v),T[i].splay(v);
					if(T[i].lc(v)!=u||T[i].rc(u))continue;
					dn=0;
					if(i==w)puts("Success.");
					else if(d[w][u]>=2||d[w][v]>=2)puts("Error 1.");
					else if(T[w].findroot(u)==T[w].findroot(v))puts("Error 2.");
					else
					{
						T[i].cut(u,v);T[w].link(u,v);
						d[i][u]--;d[i][v]--;
						d[w][u]++;d[w][v]++;
						puts("Success.");
					}
					break;
				}
			if(dn)puts("No such edge.");
		}
		if(opt==2)
		{
			int c,u,v;
			read(c);read(u);read(v);
			c++;
			if(T[c].findroot(u)!=T[c].findroot(v))puts("-1");
			else T[c].split(u,v),write(T[c].Mx[v],'\n');
		}
	}
	return 0;
}